1147 Heaps

Posted 2020-12-31 kkmjy

1147 Heaps（30 分）

In computer science, a heap is a specialized tree-based data structure that satisfies the heap property: if P is a parent node of C, then the key (the value) of P is either greater than or equal to (in a max heap) or less than or equal to (in a min heap) the key of C. A common implementation of a heap is the binary heap, in which the tree is a complete binary tree. (Quoted from Wikipedia at https://en.wikipedia.org/wiki/Heap_(data_structure))

Your job is to tell if a given complete binary tree is a heap.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers: M (≤ 100), the number of trees to be tested; and N (1 < N ≤ 1,000), the number of keys in each tree, respectively. Then M lines follow, each contains N distinct integer keys (all in the range of int), which gives the level order traversal sequence of a complete binary tree.

Output Specification:

For each given tree, print in a line Max Heap if it is a max heap, or Min Heap for a min heap, or Not Heap if it is not a heap at all. Then in the next line print the tree‘s postorder traversal sequence. All the numbers are separated by a space, and there must no extra space at the beginning or the end of the line.

Sample Input:

3 8

98 72 86 60 65 12 23 50

8 38 25 58 52 82 70 60

10 28 15 12 34 9 8 56

Sample Output:

Max Heap

50 60 65 72 12 23 86 98

Min Heap

60 58 52 38 82 70 25 8

Not Heap

56 12 34 28 9 8 15 10

 

题意：

给出若干个完全二叉树的层序遍历，判断该二叉树是否满足大顶堆/小顶堆的性质，并输出后序序列。 

 

思路：

利用完全二叉树顺序存储时的性质，都不用建树，直接对数组进行操作即可。层序遍历二叉树（也就是顺序遍历数组）的非叶结点，逐个判断。

 

代码：

#include <cstdio>

int queryCnt,n;
int CBT[1005];

void postOrderTraversal(int root)
{
    if(root>n) return;
    postOrderTraversal(root*2);
    postOrderTraversal(root*2+1);
    printf("%d",CBT[root]);
    if(root==1) printf("
");//因为是后序遍历，最后访问根节点，而根节点的下标固定为1
    else printf(" ");
}

int main()
{
    scanf("%d%d",&queryCnt,&n);
    for(int i=0;i<queryCnt;i++){
        for(int j=1;j<=n;j++)
            scanf("%d",&CBT[j]);
        //判断，层序遍历完全二叉树的非叶结点
        int flag=CBT[1]>CBT[2]?1:0;//大顶堆标记为1，小顶堆标记为0。题目保证结点个数至少有两个。
        for(int j=1;j<=n/2;j++){
            if(flag==1){
                if(CBT[j]<CBT[j*2] || (j*2+1<=n && CBT[j]<CBT[j*2+1])){
                    flag=-1;
                    break;
                }
            }
            if(flag==0){
                if(CBT[j]>CBT[j*2] || (j*2+1<=n && CBT[j]>CBT[j*2+1])){
                    flag=-1;
                    break;
                }
            }
        }
        if(flag==1) printf("Max Heap
");
        else if(flag==0) printf("Min Heap
");
        else printf("Not Heap
");
        //输出后续序列
        postOrderTraversal(1);
    }
    return 0;
}